Xias wrote:I could have defined it to include the endpoints. Then at midnight, there would be exactly one element in the set, and that element would be zero.

It would have one element because set theory coalesced an infinite number of zero values into one. That is an indication that set theory is not the right framework for that particular problem - it's thrown away information. If you modeled the numbers as a series then you would indeed end up with 0,0,0,... Or if you modeled each element as a pair of values including the n that generated it, you would end up with an infinite set {{0,1},{0,2},{0,3},...}

poker wrote:What happens when x is infinite? First, x cannot be part of Bj for any finite j, since since each finite Bj only contains finite values.

phlip wrote:So now, we need to do the same here... in order for L_inf to be non-empty there needs to be some specific I_j which is non-empty.

Xias wrote:What stops me from arguing the same way for {n+1,...10n} at the limit? Nothing! I am arguing the same way for {n+1,...10n} at the limit!

Well, if we're considering looking at the limit (which we must if we are to get past midnight):

Defining Bn={n+1,n+2,...10n}

Lsup = ∩n≥1∪j≥nBjEvery ∪j will contain all members of B∞, therefore the intersection of all ∪j also contains all members of B∞

Linf = ∪n≥1∩j≥nBj∩j=∞Bj = B∞, therefore the union of all ∩j≥nBj = B∞

So the limit of B is B∞ and I can't see any way to claim that it's empty.

Xias wrote:It is especially ironic for you to argue this, since the entire point of the puzzle is to reference a particular solution to the Ross-Littlewood problem. If we add ten balls and remove one ball at every step, the jug will intuitively be full at midnight. However, if we remove balls in a particular and systematic way, we end up with an empty jug, leading to the Ross-Littlewood Paradox. This is the essence of the puzzle.

The paradox is that there are a number of rational arguments for there being infinite infinitely-numbered balls in the jug. These arguments seem to be contradicted by the statement S: "For every number that enters the jug I can give you a time when it leaves the jug". That is what seems at first glance to be paradoxical - but the statement S is only true for finite numbers, so it does not in fact lead to a paradox. Every finite number leaves the jug but actually at midnight an infinite number are added and an infinite number leave - and that does not imply that nothing is left, since we know that infinity minus infinity cannot be said to be equal to zero.

The set-theory argument using strictly finite numbers is simply a mathematical formulation of the initial statement S. By ignoring infinite numbers you make it seem as though nothing is left, whereas in fact all you can say is that no finite numbers are left. If you take the limit actually to infinity, not just approaching it, then the set B∞ (an infinite set of infinite numbers) remains and the paradox is resolved.

poker wrote:What happens when x is infinite? First, x cannot be part of Bj for any finite j, since since each finite Bj only contains finite values.

phlip wrote:So now, we need to do the same here... in order for L_inf to be non-empty there needs to be some specific I_j which is non-empty.

Xias wrote:What stops me from arguing the same way for {n+1,...10n} at the limit? Nothing! I am arguing the same way for {n+1,...10n} at the limit!

Well, if we're considering looking at the limit (which we must if we are to get past midnight):

Defining Bn={n+1,n+2,...10n}

Lsup = ∩n≥1∪j≥nBjEvery ∪j will contain all members of B∞, therefore the intersection of all ∪j also contains all members of B∞

Linf = ∪n≥1∩j≥nBj∩j=∞Bj = B∞, therefore the union of all ∩j≥nBj = B∞

So the limit of B is B∞ and I can't see any way to claim that it's empty.

I see. So back when I said:

Poker wrote:Therefore, if you want to claim that x is in the lim_sup, you must also claim that the process used to define lim_sup must itself use a set of numbers that extends beyond the natural numbers. Is this something you are prepared to claim?

And:

Poker wrote:Therefore, if you want to claim that x is in the lim_inf, you must also claim that the process used to define lim_inf must itself use a set of numbers that extends beyond the natural numbers. Is this something you are prepared to claim?

Your answer to that question is apparently yes? Remember, I'm not talking about the balls here - I'm talking about the method used to define limits themselves. What justification do you have for doing this?

If you're going to be so careless with the notation, it's difficult to really address what you are saying. When we have a ∪n≥1 (or ∩n≥1), what we are saying is "evaluate what is on the right for every value of n greater than or equal to 1, then take the union (or intersection) of every result."

Then what is on the right is ∩j≥n (or ∪ j≥n), which is saying "evaluate what is on the right for every value of j greater than or equal to n, then take the intersection (or union) of every result." In this case, the value of n is dictated by what the input is from the ∪n≥1 (or ∩n≥1) on the left.

Part of the reason why you run into this problem:

kryptonaut wrote:So the limit of B is B∞ and I can't see any way to claim that it's empty.

is that you didn't actually do any evaluating with the specific sequence that you defined. The result that "the limit of B is B∞" is unsurprising, then, since you only worked with the generality. So let's actually do the evaluating with the specific sequence you presented.

With limsup, we have Lsup = ∩n≥1∪j≥nBj. The leftmost part says "evaluate what is on the right for every n greater than or equal to 1, then take the intersection of every result." So let's start by evaluating what is on the right for n=1. - We have ∪j≥1Bj. The leftmost part says "evaluate what is on the right for every j greater than or equal to 1, then take the union of every result." So let's start by evaluating what is on the right for j=1.-- We have B1 = {2, 3, 4, 5, 6, 7, 8, 9, 10}.- Then let's evaluate what is on the right for j=2.-- We have B2 = {3, 4, ... , 19, 20}.- We can continue this for every j≥1. Then we take the union of every result, and get B1 ∪ B2 ∪ B3 ∪ ... = {2, 3, 4, ...}Then we repeat this evaluation for n=2:- We have ∪j≥2Bj. The leftmost part says "evaluate what is on the right for every j greater than or equal to 2, then take the union of every result." So let's start by evaluating what is on the right for j=2.-- We have B2 = {3, 4, ... , 19, 20}.- Then let's evaluate what is on the right for j=3.-- We have B3 = {4, 5, ... , 29, 30}.- We can continue this for every j≥2. Then we take the union of every result, and get B2 ∪ B3 ∪ B4 ∪ ... = {3, 4, 5, ...}We can continue this for every n≥1. Then we take the intersection of every result, and get ∪j≥1Bj∩ ∪j≥2Bj∩ ∪j≥3Bj∩ ... = {2, 3, 4, ...} ∩ {3, 4, 5, ...} ∩ {4, 5, 6, ...} ∩ ... which is equal to the null set.

With liminf, we have ∪n≥1∩j≥nBj. The leftmost part says "evaluate what is on the right for every n greater than or equal to 1, then take the union of every result." So let's start by evaluating what is on the right for n=1. - We have ∩j≥1Bj. The leftmost part says "evaluate what is on the right for every j greater than or equal to 1, then take the intersection of every result." So let's start by evaluating what is on the right for j=1.-- We have B1 = {2, 3, 4, 5, 6, 7, 8, 9, 10}.- Then let's evaluate what is on the right for j=2.-- We have B2 = {3, 4, ... , 19, 20}.- We can continue this for every j≥1. Then we take the intersection of every result, and get B1 ∩ B2 ∩ B3 ∩ ... = {}. Then we repeat this evaluation for n=2:- We have ∩j≥2Bj. The leftmost part says "evaluate what is on the right for every j greater than or equal to 2, then take the intersection of every result." So let's start by evaluating what is on the right for j=2.-- We have B2 = {3, 4, ... , 19, 20}.- Then let's evaluate what is on the right for j=3.-- We have B3 = {4, 5, ... , 29, 30}.- We can continue this for every j≥2. Then we take the intersection of every result, and get B2 ∩ B3 ∩ B4 ∩ ... = {}We can continue this for every n≥1. Then we take the union of every result, and get ∩j≥1Bj∪ ∩j≥2Bj∪ ∩j≥3Bj∪ ... = {} ∪ {} ∪ {} ∪ ... which is equal to the null set.

Edit: For the sake of illustration, I'll show this process when applied to the "remove the lowest even numbered ball" game:

Spoiler:

With limsup, we have Lsup = ∩n≥1∪j≥nBj. The leftmost part says "evaluate what is on the right for every n greater than or equal to 1, then take the intersection of every result." So let's start by evaluating what is on the right for n=1. - We have ∪j≥1Bj. The leftmost part says "evaluate what is on the right for every j greater than or equal to 1, then take the union of every result." So let's start by evaluating what is on the right for j=1.-- We have B1 = {1, 3, 4, 5, 6, 7, 8, 9, 10}.- Then let's evaluate what is on the right for j=2.-- We have B2 = {1, 3, 5, 6, 7, ... , 19, 20}.- We can continue this for every j≥1. Then we take the union of every result, and get B1 ∪ B2 ∪ B3 ∪ ... = {1, 3, 4, 5, 6, 7, ...} or every natural number except for 2.Then we repeat this evaluation for n=2:- We have ∪j≥2Bj. The leftmost part says "evaluate what is on the right for every j greater than or equal to 2, then take the union of every result." So let's start by evaluating what is on the right for j=2.-- We have B2 = {1, 3, 5, 6, 7, ... , 19, 20}.- Then let's evaluate what is on the right for j=3.-- We have B3 = {1, 3, 5, 7, 8, 9, ... , 29, 30}.- We can continue this for every j≥2. Then we take the union of every result, and get B2 ∪ B3 ∪ B4 ∪ ... = {1, 3, 5, 6, 7, 8, 9, ...} or every natural number except for 2 and 4.We can continue this for every n≥1. Then we take the intersection of every result, and get ∪j≥1Bj∩ ∪j≥2Bj∩ ∪j≥3Bj∩ ... = {1, 3, 4, 5, 6, 7, ...} ∩ {1, 3, 5, 6, 7, 8, 9, ...} ∩ {1, 3, 5, 7, 8, 9, 10, 11, ...} ∩ ... which is equal to {1, 3, 5, 7, ...} or the set of all odd numbers.

With liminf, we have ∪n≥1∩j≥nBj. The leftmost part says "evaluate what is on the right for every n greater than or equal to 1, then take the union of every result." So let's start by evaluating what is on the right for n=1. - We have ∩j≥1Bj. The leftmost part says "evaluate what is on the right for every j greater than or equal to 1, then take the intersection of every result." So let's start by evaluating what is on the right for j=1.-- We have B1 = {1, 3, 4, 5, 6, 7, 8, 9, 10}.- Then let's evaluate what is on the right for j=2.-- We have B2 = {1, 3, 5, 6, 7, ... , 19, 20}.- We can continue this for every j≥1. Then we take the intersection of every result, and get B1 ∩ B2 ∩ B3 ∩ ... = {1, 3, 5, 7, ...} or every odd number.Then we repeat this evaluation for n=2:- We have ∩j≥2Bj. The leftmost part says "evaluate what is on the right for every j greater than or equal to 2, then take the intersection of every result." So let's start by evaluating what is on the right for j=2.-- We have B2 = {1, 3, 5, 6, 7, ... , 19, 20}.- Then let's evaluate what is on the right for j=3.-- We have B3 = {1, 3, 5, 7, 8, 9, ... , 29, 30}..- We can continue this for every j≥2. Then we take the intersection of every result, and get B2 ∩ B3 ∩ B4 ∩ ... = {1, 3, 5, 7, ...} or every odd number.We can continue this for every n≥1. Then we take the union of every result, and get ∩j≥1Bj∪ ∩j≥2Bj∪ ∩j≥3Bj∪ ... = {1, 3, 5, 7, ...} ∪ {1, 3, 5, 7, ...} ∪ {1, 3, 5, 7, ...} ∪ ... which is equal to {1, 3, 5, 7, ...} or the set of all odd numbers.

This method is valid for calculating the limit of any sequence of sets, so long as liminf and limsup are the same.

kryptonaut wrote:Every finite number leaves the jug but actually at midnight an infinite number are added and an infinite number leave - and that does not imply that nothing is left, since we know that infinity minus infinity cannot be said to be equal to zero.

Tell me, kryptonaut, where on earth did you get that we were doing anything at midnight at all, if not from your own imagination? The puzzle certainly doesn't prescribe it. When we only do things as prescribed by the puzzle - that is, we only operate during the infinitely many finite numbered steps before midnight - there is only one reasonable conclusion.

Let's pretend there is a midnight step. Take whatever you are imagining happens at midnight: in your mind, an infinite number of balls are added, and an infinite number of balls are removed. Let's walk through the "step:"

1. None of the balls in the jug before the step are infinitely numbered, since we have only had finitely numbered steps up to this point.2. None of the balls you add at midnight are finitely numbered, because every finitely numbered ball is added at some step before midnight.3. None of the balls you remove at midnight are finitely numbered, because every finitely numbered ball is removed at some step before midnight.4. After the step, there are no finitely numbered balls in the jug.

So after the midnight step, despite not removing any finitely numbered balls during the midnight step, there are no finitely numbered balls left. So there were no finitely numbered balls in the jug before the step at midnight. And there were no infinitely numbered balls in the jug before the step at midnight. So before you do anything during the midnight step, the jug must be empty since nothing can be in it - infinite or not.

So these infinitely numbered balls that you claim are in the jug are merely the product of performing some action at a ωth step. Before that step, the jug is empty. So if you do nothing during the ωth step, the jug remains empty. If you stow away ucim's long-lost ham sandwiches in the jug during the ωth step, then the jug will be full of ham sandwiches.

Now where in the original puzzle is it prescribed that we do anything at all in this ωth step, let alone specifically adding infinitely numbered balls, let alone an infinite number of them?

Lsup = ∩n≥1∪j≥nBjEvery ∪j will contain all members of B∞, therefore the intersection of all ∪j also contains all members of B∞

Linf = ∪n≥1∩j≥nBj∩j=∞Bj = B∞, therefore the union of all ∩j≥nBj = B∞

So the limit of B is B∞ and I can't see any way to claim that it's empty.

But... B∞ isn't a thing that exists. Bn has been defined for all natural n, all of which are finite.

∪j≥nBj, for each n, is a union of Bj for all natural j greater than or equal to n. It doesn't "contain all members of B∞" because that would require that (a) ∞ be a natural number, which it's not, and (b) B∞ be a thing that exists, which it doesn't.

Similarly, ∩j≥nBj cannot be evaluated at n=∞, because it is only defined for natural n, and ∞ is not natural.

Or... at least... that's my answer given what I think you're saying, given your abuse of the notation. Like Xias said, you continue to play fast and loose with the notation, which makes it hard sometimes to understand what you're actually trying to say. Like... I interpreted this:

kryptonaut wrote:Every ∪j will contain all members of B∞

to mean this

kryptonaut, translated wrote:Every ∪j≥nBj will contain all members of B∞ because I mistakenly believe that this j is able to be infinite, and ∞≥n

because at least that makes some amount of sense, even if it's incorrect. But on the other hand, it's also possible that you think literally "∪j" is an object in and of itself, and are completely misunderstanding what the notation even means.

This is why it's important to be precise in what you're saying, especially when you have such a fundamental disagreement like this. It's important that you're very clear with exactly what you're arguing, so that the discussion can stay on track, and not get diverged into people arguing against things they think you're saying, but you don't actually mean... that doesn't help anyone.

I've been thinking some more about this (don't groan ), particularly in light of mward's post.

I imagined an infinitely long train with numbered carriages, passing through a station. At any finite time it's possible to say which carriage is passing through - but after an infinite number of carriages have passed through, it is not. But that doesn't mean there is no carriage, just that we don't know it's number. All we can say is that an infinite number of carriages passed through the station before it. This means that the current carriage has ordinal ω in the train. ω does not correspond to a particular cardinal number, it just means that a countable infinite number of other ordinals come before it.

If the station platform was long enough for, say, 8 carriages then we could say the first one had ordinal ω, the second ω+1, up to ω+7. We'd have no information about the actual cardinals involved; they never 'settle down' as the train keeps moving.

This is analogous to ucim's 'holding jug' scenario, in which an infinite never-ending stream of balls passes through the holding jug C into a discard pile. C always contains a single ball, ultimately with ordinal ω when there are an infinite number of balls in the discard pile. We don't know the cardinal number of the ball in C, it never 'settles down'. More balls keep passing through the holding jug. The one in it always has ordinal ω since there is always an infinite set in the discard pile.

So there is no 'number' that we can say ends up in C; no numbered carriage we can say ends up at the station, but we can still say there is 1 ball in C or 8 carriages passing through the station, and we can say that an infinite number have gone before them.

In the same way, in the 'remove lowest' puzzle, the jug never settles down to a fixed set of cardinals, since they all get removed some time/steps after arriving - but that doesn't mean that the jug is empty - an infinite number of balls have left it, so the ordinal of the first ball in the jug can be said to be ω, and there are an infinite number of balls in the jug, so the set is {ω,ω+1,ω+2,...} - it's just that the actual cardinal numbers are unknowable.

Now, there was no 'ω-numbered' ball in the input set - but since this is an infinite set it is possible to take an infinite subset of it (the discarded balls), and what comes after that subset does have ordinal ω in that partitioning. The cardinals are still sequential from the first discarded ball (1) through the first ball in the jug (unknowable) and up to the last in the jug (also unknowable), so the union of the set of cardinals of discarded balls and those in the jug still equates to the input set. I guess there are always more to come, and we'll never get to a ball with cardinal Aleph-null.

kryptonaut wrote:I've been thinking some more about this (don't groan ), particularly in light of mward's post.

I imagined an infinitely long train with numbered carriages, passing through a station. At any finite time it's possible to say which carriage is passing through - but after an infinite number of carriages have passed through, it is not. But that doesn't mean there is no carriage, just that we don't know it's number. All we can say is that an infinite number of carriages passed through the station before it. This means that the current carriage has ordinal ω in the train. ω does not correspond to a particular cardinal number, it just means that a countable infinite number of other ordinals come before it.

If the station platform was long enough for, say, 8 carriages then we could say the first one had ordinal ω, the second ω+1, up to ω+7. We'd have no information about the actual cardinals involved; they never 'settle down' as the train keeps moving.

I'm going to join in on the "asking you to be a bit more precise" bandwagon here: exactly what set of numbers is being used to number the carriages, and exactly what ordering? If they are the natural numbers in their usual order, then after an infinite number of carriages have passed through, there are no carriages left to pass through. If they were ordinal numbers, which include ω (as well as ω+1, ω+2, ..., 2ω, 3ω, and oh so far beyond), then what you are saying could have some merit, if only because simply saying "infinite" does not tell us exactly which ordinal version of infinite have passed through (yet another example of not being precise enough).

Also, what exactly do you mean by "cardinals"? Every ordinal you've listed, and also that I have specifically mentioned, has the same cardinal value: aleph-null.

kryptonaut wrote:This is analogous to ucim's 'holding jug' scenario, in which an infinite never-ending stream of balls passes through the holding jug C into a discard pile. C always contains a single ball, ultimately with ordinal ω when there are an infinite number of balls in the discard pile. We don't know the cardinal number of the ball in C, it never 'settles down'. More balls keep passing through the holding jug. The one in it always has ordinal ω since there is always an infinite set in the discard pile.

So there is no 'number' that we can say ends up in C; no numbered carriage we can say ends up at the station, but we can still say there is 1 ball in C or 8 carriages passing through the station, and we can say that an infinite number have gone before them.

What you are saying only makes some sort of sense if we are dealing with a set for which there is an infinite number of values after an infinite number of values. You can certainly invent orderings of the natural numbers for which such a thing is true (the first-all-odds-then-all-evens ordering, for example), but what proof do you have that you can do this for this particular set and this particular ordering?

Actually, looking at this more carefully, I think you somehow have ordinals and cardinals reversed here. Cardinals deal with amounts; ordinals deal with the order. After an infinite number of balls are in the discard pile, if there is something else, its cardinal value is aleph-null. The ordinal value, immediately after the infinite number of balls have "passed through", would be ω, but if it doesn't "settle down" the ordinal would change, to ω+1, then ω+2, then ω+3, and so on, as ordinals do.

kryptonaut wrote:In the same way, in the 'remove lowest' puzzle, the jug never settles down to a fixed set of cardinals, since they all get removed some time/steps after arriving - but that doesn't mean that the jug is empty - an infinite number of balls have left it, so the ordinal of the first ball in the jug can be said to be ω, and there are an infinite number of balls in the jug, so the set is {ω,ω+1,ω+2,...} - it's just that the actual cardinal numbers are unknowable.

Again, you casually say "the first ball in the jug" as though it is a given that it exists. And again, I'm not sure what you mean by "cardinals". If you mean to say that the "cardinal" usual values are different from the "ordinal" order values (which is just my guess about what you mean, since, again, you're not being very precise), that notion is mistaken right from the get-go, because the order in which the balls are put into, and removed from, the jug is the same as the usual order of the natural numbers - that is, the "cardinal" values and the "ordinal" values are exactly the same!

kryptonaut wrote:Now, there was no 'ω-numbered' ball in the input set - but since this is an infinite set it is possible to take an infinite subset of it (the discarded balls), and what comes after that subset does have ordinal ω in that partitioning. The cardinals are still sequential from the first discarded ball (1) through the first ball in the jug (unknowable) and up to the last in the jug (also unknowable), so the union of the set of cardinals of discarded balls and those in the jug still equates to the input set.

In what partitioning, exactly? The discarded balls vs the balls remaining in the jug at some particular time you're imagining such that the number of balls is infinte? (Again, this is only my guess about what you mean.) Again, you would need a number that comes after an infinite number of numbers, in their usual order. And again, this is not possible in the natural numbers.

kryptonaut wrote:I guess there are always more to come, and we'll never get to a ball with cardinal Aleph-null.

I was about to say something about how if you haven't gotten to aleph-null yet then you've only gone through finitely-many balls, but that would be making a huge assumption about what you mean by "cardinal" numbers.

Poker wrote:If they were ordinal numbers, which include ω (as well as ω+1, ω+2, ..., 2ω, 3ω, and oh so far beyond), then what you are saying could have some merit, if only because simply saying "infinite" does not tell us exactly which ordinal version of infinite have passed through (yet another example of not being precise enough).

It's like saying, "It's possible to say which car is going through after 10 seconds, but it's not possible to say which car is passing through after an odd number of seconds."

Sure, that's true, but it's not because there's something wrong or weird about the idea of an odd number, but rather because "an odd number of seconds" simply isn't a precise amount of time.

Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.---If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

Surely midnight can't arrive, because the person moving these balls around has to move faster and faster to do so and will thus experience time more and more slowly.

Also, if the distance between "in the jug" and "out of the jug" is a piffling 1cm, by the time you've only got 1/2256s available in which to move the balls in and one of them out you need to move them at an average speed of 1/(100*2512) = 1.34*10152m/s ... and at that velocity, er, what's the formula for kinetic energy in superluminal projectiles again? Heck, even before that, 0.99999c gives kinetic energy of nearly 4.5*1021 J/kg, so more than 104 times the energy density of TNT.

Ever been asked to define 1? I can define 1. 1 is the average number of new patients each infected person has to infect for an epidemic to be neither fading away nor getting worse. This definition is no use whatsoever on a mathematical degree course, and Xeno's paradox is no use whatsoever off one.

Sableagle wrote:Surely midnight can't arrive, because the person moving these balls around has to move faster and faster to do so and will thus experience time more and more slowly.

Also, if the distance between "in the jug" and "out of the jug" is a piffling 1cm, by the time you've only got 1/2256s available in which to move the balls in and one of them out you need to move them at an average speed of 1/(100*2512) = 1.34*10152m/s ... and at that velocity, er, what's the formula for kinetic energy in superluminal projectiles again? Heck, even before that, 0.99999c gives kinetic energy of nearly 4.5*1021 J/kg, so more than 104 times the energy density of TNT.

Ever been asked to define 1? I can define 1. 1 is the average number of new patients each infected person has to infect for an epidemic to be neither fading away nor getting worse. This definition is no use whatsoever on a mathematical degree course, and Xeno's paradox is no use whatsoever off one.

I have several different responses, depending on the nature of... well, you, I guess:

[fakeanger]How DARE you try to bring realism into this![/fakeanger]

[retort]Given the nature of the problem, particularly the doubling-speed-infinitely part, I am more inclined to believe this is on said course, rather than off.[/retort]

[probablylowqualitywit]Well, you've got one relevant line here: "What we've actually demonstrated here is that humanity can't process infinity." Given the length of this thread, this is largely true.[/probablylowqualitywit]

kryptonaut wrote:This is analogous to ucim's 'holding jug' scenario, in which an infinite never-ending stream of balls passes through the holding jug C into a discard pile. C always contains a single ball

No, C doesn't always contain a single ball. Half the time it's empty. We put a ball into C, we take a ball out of C, we put a ball into C, we take a ball out of C...

Now, in a regular (infinite) task, we never finish. But in a supertask we do, because each iteration happens faster and faster.

How much time does ball n spend in C? Let n increase without bound; the amount of time ball n spends in C decreases towards zero. When n "equals" infinity, the amount of time that respective ball spends in C "equals" zero. Therefore the jug is empty at the end of the supertask.

Suppose there were a ball labeled ω in C at the end of the supertask? What was the label on the ball that came just before it? There is no such ball, and there is no such label, because there is no number that is "closest" to (t=)1 (or zero, or any real number for that matter). The ball labeled ω cannot be put into C until the last natural numbered ball gets removed. That never happens.

This is the key. Even though time moves past t=1 (or whatever the finish point is), and every natural numbered ball has gotten removed, there is no last natural numbered ball to get removed to make way for the putative ball labeled ω. Therefore the jug is empty at the end of the supertask.

kryptonaut wrote:The one in it always has ordinal ω since there is always an infinite set in the discard pile.

The one in it has a label with the natural number ω. However, there is no such number. Therefore the jug is empty at the end of the supertask.

kryptonaut wrote:So there is no 'number' that we can say ends up in C

Therefore the jug is empty at the end of the supertask.

Jose

Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

kryptonaut wrote:This is analogous to ucim's 'holding jug' scenario, in which an infinite never-ending stream of balls passes through the holding jug C into a discard pile. C always contains a single ball

No, C doesn't always contain a single ball. Half the time it's empty. We put a ball into C, we take a ball out of C, we put a ball into C, we take a ball out of C...

Does that matter? I mean your answer probably wouldn't change if it was specified that you use the new ball to push the old one out?

We could say anything, really. What matters is what we can justify. So when you say this, what you're actually saying is "then I am going to say..." unless you actually justify it.

kryptonaut wrote:it's just that the actual cardinal numbers are unknowable.

What we have is a situation where you are assuming that there is a ball in the jug. Then the number on that jug being a natural number leads to a contradiction, so it must be nonnatural. But it being a nonnatural number leads to a contradiction, so it must be natural. This is in and of itself a contradiction, so the original assumption must be wrong and there must be no ball in the jug.

So are you saying that it is unknowable because it's actually unknowable, or because you personally don't know how to justify it without leading to a contradiction?

kryptonaut wrote:but since this is an infinite set it is possible to take an infinite subset of it (the discarded balls), and what comes after that subset does have ordinal ω in that partitioning.

Are you just ignoring the multiple explanations of partitions that have been offered to you? You can't partition N in that way. The only subset of N that is ordered sequentially from 1 and has infinite elements is N itself. In fact, that's pretty much the definition of N.

I will point you to this one more time:

Xias wrote:Let's pretend there is a midnight step. Take whatever you are imagining happens at midnight: in your mind, an infinite number of balls are added, and an infinite number of balls are removed. Let's walk through the "step:"

1. None of the balls in the jug before the step are infinitely numbered, since we have only had finitely numbered steps up to this point.2. None of the balls you add at midnight are finitely numbered, because every finitely numbered ball is added at some step before midnight.3. None of the balls you remove at midnight are finitely numbered, because every finitely numbered ball is removed at some step before midnight.4. After the step, there are no finitely numbered balls in the jug.

So after the midnight step, despite not removing any finitely numbered balls during the midnight step, there are no finitely numbered balls left. So there were no finitely numbered balls in the jug before the step at midnight. And there were no infinitely numbered balls in the jug before the step at midnight. So before you do anything during the midnight step, the jug must be empty since nothing can be in it - infinite or not.

kryptonaut wrote:This is analogous to ucim's 'holding jug' scenario, in which an infinite never-ending stream of balls passes through the holding jug C into a discard pile. C always contains a single ball

No, C doesn't always contain a single ball. Half the time it's empty. We put a ball into C, we take a ball out of C, we put a ball into C, we take a ball out of C...

Does that matter? I mean your answer probably wouldn't change if it was specified that you use the new ball to push the old one out?

No, it doesn't really matter, but I was trying to offer kryptonaut a bit of intuition to counter xis intuition that the jug has to have something in it. I was leading up to the idea that the amount of time a ball spends in the jug tends towards zero (which is of course true anyway). I think xis issue is that it "doesn't make sense", not that it "can't be proven". So I'm offering a way for it to "make sense" in his mind, so that xe can then be open to accepting the other rigorous proofs that have been offered.

Jose

Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

mward wrote:I think that this method can be applied to any supertask: and will either calculate the result, or will prove that the result is indeterminate. I would love to be proved wrong, however, so if you have a counterexample, please post it!

I like your reasoning, particularly as you seem prepared to at least admit to the presence of ordinals greater than or equal to ω .Clearly I disagree with your conclusion in case (1), the original 'remove-the-lowest' puzzle. My reasoning, in your framework, is that although at the end of the task ω balls have been added and removed, for each ball removed 9 were added and never removed, leaving 9ω (which equals ω) behind.

In my reasoning, any ball which is left behind must have been left behind by a finite subtask.Which finite subtask left any of these ω balls in the jug?

Note: ω is an ordinal number. The ordinal numbers are the arrangements of the elements of a set in a particular order (first, second, third etc). ω is the smallest infinite ordinal number. The cardinal numbers are the number of elements in a set, aleph null (ℵ0) is the smallest infinite cardinal number.

To simplify the argument, consider a jug which contains all the (finite) natural numbered balls and nothing else. There are ℵ0 balls in the jug, numbered 1, 2, 3, ... up to, but not including ball ω.

I will describe three simple "supertasks" (A, B and C) starting with this jug:(A) One step one: take all balls out of the jug. For steps 1 < n < ω, do nothing.Are we agreed that at midnight the jug is empty?We can split this supertask into ω finite tasks, TA1, TA2, ... (up to but not including ω) where TAn is the task:

Step 1: Take ball n from the jug

Step n, for 1 < n < ω: Do nothing

Are we agreed that the supertask can be split up into these finite tasks which all run in parallel and do not interact with each other?The number of balls in the jug on each step is: ℵ0, 0, 0, 0, ...The number of balls in the jug at midnight is 0.

(B) One step n, for 1 <= n < ω take ball n from the jug.We can split this supertask into ω finite tasks, TB1, TB2, ... (up to but not including ω) where TBn is the task:

Step n: Take ball n from the jug

All other steps: Do nothing

For every ball n, the finite task TBn takes ball n from the jug and does not put it back. In fact, TBn does the same as TAn, but does it a finite time later. So the overall effect of TBn on the jug at midnight is identical to the effect of TAn on the jug at midnight.So, at midnight after task B the jug is empty: the same as for A.

Are we agreed?

The set of balls in the jug at step n is {n, n+1, ...}. This set has cardinality ℵ0.So the number of balls in the jug at each step is precisely:

ℵ0, ℵ0, ℵ0, ℵ0, ℵ0, ℵ0, ℵ0, ...

The "limit" of this sequence of cardinal numbers is: ℵ0.But the number of balls in the jug at midnight is zero.So the limit of the sequence of numbers of balls in the jug before midnight is NOT equal to the number of balls in the jug at midnight!

Are we agreed?

So, and this is crucial, we should not expect the limit of the number of balls in the jug before midnight to bear any relation to the number of balls in the jug at midnight. We can get the same sequence of "number of balls in the jug" by doing nothing on each step: in which case all the balls are still in the jug at midnight.

So, given only the sequence of (cardinal) numbers telling us how many balls are in the jug on each step, we cannot determine the number of balls in the jug at midnight. This number can be calculated by determining the effect on the jug of each finite subtask (which balls are left in the jug, or placed in the jug and not removed, by each subtask) and combining the results. Since the subtasks are all independent, we can combine the results for each subtask to get the result of the supertask (the actual set of balls in the jug at midnight), and then examine this set to count the number of balls in the jug at midnight.

(C) For task C, imagine that the jug is divided into two halves, L and R, and initially all the balls are in the L half. The finite subtasks are TCn:

On step int((n-1)/10)+1 move ball n from L to R.

On step n take ball n (which will be in side R) out of the jug

All other steps: do nothing

As far as the jug as a whole is concerned, this task is identical to task B. TCn takes ball n out of the jug and never puts it back. The rest of task TCn is shuffling balls around within the jug. As far as the state of the jug at midnight is concerned, task C does the same as task B which does the same as task A but with a delay for each ball.

So, at midnight after task C the jug is empty.

Are we agreed?

If the jug is empty at midnight, then clearly both halves of the jug L and R are empty.

But now, if we focus on the R half of the jug we see that the movement of balls into and out of this half of the jug is precisely the same as the movement of balls into and out of the jug in the original problem!

Therefore, in the original problem, the jug is empty at midnight. QED.

mward wrote:We can split this supertask into ω finite tasks, TA1, TA2, ... (up to but not including ω) where TAn is the task:

Do you perhaps mean "We can split this supertask into ℵ0 finite tasks, TA1, TA2, ... (up to but not including ω) where TAn is the task:"?

(Just a nit, but it's the same nit you picked in your second paragraph. )

Jose

Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

Yes, the ordinal number is appropriate for the order (TAω, which never happens). But the number of tasks is still a cardinal number: ℵ0.

Or am I missing a subtlety?

Jose

Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

I believe what he meant was 'we can split this supertask into a well-ordered set of finite tasks isomorphic to omega, each element denoted TAn for some finite ordinal n', at least that's what I got from it.

Cauchy wrote:At the very least, I think it's clear that this is a matter of opinion now, not math. You can clearly take the limits of the p and f functions philip outlined, and they have the obvious limits. Whether you think p or f describes the supertask better is not something that can be deduced, really.

It can't be deduced purely mathematically, but it can still be argued on the basis of what you want to be able to say is true about supertasks.

As Xias already mentioned, two balls both labeled "1" are still two balls, unless you define "ball" differently from anyone else, irrespective of whether one of them "mystically" remembers what label it had before.

In your opinion, what happens if at step 1 we add a ball with the label "1", and then at step n>1 we change the label on the ball to "n"? Is the jug empty at midnight? If the fact that no natural number label can be on the ball implies that the ball must have left the jug, why doesn't the fact that no ball can be in the jug (in a setup of your choice that you believe ends with that result) imply that the jug itself must have disappeared? If balls are identified entirely by what label they have, why are jugs not identified entirely by what balls they have in them?

ucim wrote:Do you perhaps mean "We can split this supertask into ℵ0 finite tasks, TA1, TA2, ... (up to but not including ω) where TAn is the task:"?

The tasks are ordered, so TA1 is the first task, TA2 is the second, and so on.So the ordinal number is appropriate.

Right, so we can describe it as an ω-sequence of tasks, but I agree with ucim that this isn't the same as using ω to describe how many tasks we have, which is grammatically what it seems like you were doing.

Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.---If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

Demki wrote:I believe what he meant was 'we can split this supertask into a well-ordered set of finite tasks isomorphic to omega, each element denoted TAn for some finite ordinal n', at least that's what I got from it.

That is indeed what I mean by "ω finite tasks, TA1, TA2, ...". I acknowledge that I appear to be just describing the number of tasks (in which case, a cardinal number is appropriate), but I am actually describing the number of tasks and the fact that they are ordered: which is why I used an ordinal number.

gmalivuk wrote:In your opinion, what happens if at step 1 we add a ball with the label "1", and then at step n>1 we change the label on the ball to "n"?

In my opinion, this case can also be dealt with by dividing supertasks into finite subtasks. Here, the operation on the ball is a supertask: the label on the ball does not "settle down" to a fixed value which remains unchanged in the rest of the process. This supertask can, however, be divided into a set of finite subtasks (a well-ordered set of finite tasks isomorphic to omega, each element denoted TLn for some finite ordinal n). The label on the ball is an ordinal number, and an ordinal number is defined as a set of ordinal numbers.

Task TL1: On step 1 add a ball with label 1 (which is the set {0} consisting of a single element, 0)

Task TLn for n>1: on step n add the finite number n-1 to the set defining the label on the ball.

We see that after step n the label on the ball is the set {0, 1, 2, ..., n-1} which is the finite number n.

At midnight the label on the ball is the set {0, 1, 2, ...} containing all the finite ordinals, which is precisely the set ω.

mward wrote:(C) For task C, imagine that the jug is divided into two halves, L and R, and initially all the balls are in the L half. The finite subtasks are TCn:On step int((n-1)/10)+1 move ball n from L to R.On step n take ball n (which will be in side R) out of the jugAll other steps: do nothing

Firstly, thanks for posting this description.

I have a problem with case C though, because of the number of iterations of each type of subtask. This is really the crux of my issue with the whole explanation - the two infinite supertasks are running in parallel but at different rates, so I don't think they are necessarily separable.

Your task C boils down to saying that on step n we move balls 10n-9...10n from L to R, and then move ball n from R out of the jug. This is essentially the same as the original problem, except everything starts out in the jug.

So when does this combined supertask end? Is it:

a) when the very last finite-numbered ball x is moved from L to R? In which case the last ball removed from the jug was numbered x/10 so there are still 9x/10 balls left (x is finite remember). Or:

b) when the last finite-numbered ball x is removed from the jug? In which case what can be said about the numbers 10x-9...10x that were transferred from L to R on the same step? Remember x is the very last finite-numbered ball. Or:

c) does the supertask never end? In which case the number of balls in L is inexhaustible, the number in R keeps increasing endlessly, and the number removed keeps increasing endlessly. That doesn't tie in with the nature of the supertask, but even if it did I can't see how it leaves R empty. It means that no particular number ends in R, but it does not mean that there are no numbers ever in R.

The way I see it, the set of balls removed from the jug is ω, meaning that the balls left in R must be numbered from ω and there are ω of them. Being an ordinal, the fact that a ball is numbered ω after the supertask does not mean it was numbered ω when it was in L. ω just means 'after a countably infinite number of steps'. In L there was no such ball. Afterwards, because of the specific way the steps were performed, there was.

This is what I mean when I say that N can be partitioned into N followed by a 'tail' of values, which are infinite in the sense that it takes an infinite number of steps to count up to them.

There is no last finite-numbered ball, and no ball with label omega (or beyond omega) can suddenly appear since at no step do we change any labels.

At midnight, there are only three options if we start with natural-numbered balls:1) Some remain in the jug. All of them have finite labels because all the balls anywhere have finite labels. 2) The jug is empty. 3) The supertask or its limit are ill-defined or nonexistent. (This is what happens if you just add and remove the same ball on successive steps.)

Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.---If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

gmalivuk wrote:There is no last finite-numbered ball, and no ball with label omega (or beyond omega) can suddenly appear since at no step do we change any labels.

At midnight, there are only three options if we start with natural-numbered balls:1) Some remain in the jug. All of them have finite labels because all the balls anywhere have finite labels. 2) The jug is empty. 3) The supertask or its limit are ill-defined or nonexistent. (This is what happens if you just add and remove the same ball on successive steps.)

Except that omega is not a cardinal label that is stuck to a particular ball, it is an ordinal, a quality of a ball that depends on where that ball is in a particular ordering. So not like '7' or '1 million' - but like 'seventh' or 'penultimate' or (in this case) 'after a countable infinite number'.

The lowest-numbered ball in R is n+1 for every step n. When n has been through every natural number, the lowest numbered ball in R is by definition given the ordinal omega.

In task C if you want to say the same set of balls is removed from the jug as is moved from L to R, then you have to demonstrate a 1:1 mapping between the two sets. Since the set removed from R is at all times a subset of the set transferred from L to R, I fail to see how this is possible.

You seem to be leaning toward my option 3, but you keep talking instead about balls and labels that simply do not and cannot exist.

If instead of sides we had colors, what do you think would happen? If each step involves removing the lowest ball and painting the next 10 balls blue, what's left at midnight?

Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.---If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

I'll reiterate: an ordinal is not a property of an element in a set. An ordinal is a set that is well ordered, and is defined in a specific way as to represent the class of sets that have an order-preserving bijection to said ordinal. When we say 'the set of balls is isomorphic to omega', we literally mean there is an order-preserving bijection between the set of balls and omega. No order-preserving operation makes omega isomorphic to any other ordinal, and every operation done in case C in mward's post is order-preserving.

You don't need to comment on this post, instead answer gmalivuk's post above.

gmalivuk wrote:You seem to be leaning toward my option 3, but you keep talking instead about balls and labels that simply do not and cannot exist.

If instead of sides we had colors, what do you think would happen? If each step involves removing the lowest ball and painting the next 10 balls blue, what's left at midnight?

I would say a set of blue balls in the jug, isomorphic to omega, and a set of blue discarded balls also isomorphic to omega, bearing numbers less than those on the balls in the jug.

If I start with an empty jug and at each step n I add two red balls numbered 2n-1 and 2n, and then paint ball n blue, so that after step n I have n blue balls {1,2,...n} followed by n red balls {n+1,n+2,...2n} - what's in the jug at the end of the supertask?

What if the balls are not numbered and I just add one red and one blue at each step?

What if I then decide to number them all, first the blue ones then the red ones?

kryptonaut wrote:I would say a set of blue balls in the jug, isomorphic to omega, and a set of blue discarded balls also isomorphic to omega, bearing numbers less than those on the balls in the jug.

An ordered set being isomorphic to ω isn't the same as telling what labels it has. The even numbers are a set of numbers order-isomorphic to ω, as are the odd numbers. Powers of 2 are a set isomorphic to ω, as are powers of 3, as are powers of every other prime, and as are the numbers that aren't prime powers. Those are just ways to partition the natural numbers into different infinite sets.

Your problem is your continued insistence that somehow balls like ω and ω+1 and 2ω just magically show up. All the balls we started with had natural numbers on them. So which natural number was originally on the now-lowest ball in the jug, which you claim is now numbered at least ω? When did it acquire its new label? How did this happen when all we've been doing is coloring some balls and removing balls one-by-one from the jug?

kryptonaut wrote:If I start with an empty jug and at each step n I add two red balls numbered 2n-1 and 2n, and then paint ball n blue, so that after step n I have n blue balls {1,2,...n} followed by n red balls {n+1,n+2,...2n} - what's in the jug at the end of the supertask?

An infinite collection of blue balls, each of which was red when it was first put into the jug but which was at some later point painted blue.

kryptonaut wrote:What if the balls are not numbered and I just add one red and one blue at each step?

Then you have an infinite number of red balls and an infinite number of blue balls.

kryptonaut wrote:What if I then decide to number them all, first the blue ones then the red ones?

Then you have completed two supertasks, each made up of ω subtasks, one after the other. If by "number them all" you mean with natural numbers, then you will have all the natural numbers on blue balls, and you will also have all the natural numbers on red balls.

If you instead choose to write some infinite ordinals on some of the balls, then instead you'd end up with some balls that have infinite ordinals on them. But this doesn't have any bearing on the tasks above, where no balls ever received infinite labels.

Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.---If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

gmalivuk wrote:In your opinion, what happens if at step 1 we add a ball with the label "1", and then at step n>1 we change the label on the ball to "n"?

There is in this case no pre-defined set of labels, so I'm less uncomfortable with emoji or ham sandwiches appearing on the balls (though in my opinion that doesn't happen). I would say that at the end, there would be a single ball whose label is infinitely long, whose first digit is "unreachable", whose last digit is unknowable, and may well be a proxy for ω (or ℵ0, depending on whether you are writing cardinals or ordinals). We would not however have a ball with a label that looks like the character "ω" or "ℵ0".

kryptonaut wrote:I have a problem with case C though, because of the number of iterations of each type of subtask. This is really the crux of my issue with the whole explanation - the two infinite supertasks are running in parallel but at different rates, so I don't think they are necessarily separable.

They may run at different rates, but they finish at the same time. That's important. I'll ask again that you consider how much time the ball spends in the transfer jug. As n goes to infinity increases without bound, the amount of time decreases towards zero.

At midnight, the amount of time that the "infinityth" ball spends in the jug is zero. The jug is empty. And it doesn't matter if it's one ball or ten balls that's being transferred in or out. So long as we keep accelerating the task so that it finishes in finite time, this remains true. The amount of time the balls spend in the jug goes to zero. Do you disagree? If so, where and why?

This would not be the case if the task were not accelerated, and therefore simply did not finish. But when forcing it, this is the inevitable result. The "last" ball spends no time in the jug.

Jose

Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

Kryptonaut, I would still like for you to address the point that I made.

Before midnight, there is only finite-numbered steps. At finite numbered step k, we only interact with the following eleven balls: {bk, b10k-9, b10k-8, b10k-7, b10k-6, b10k-5, b10k-4, b10k-3, b10k-2, b10k-1, b10k}. All of those balls have naturally numbered labels, and are indexed to a natural number (an element of the index set N.) The time associated with step k, tk, is also a time before midnight for all finite k.

You claim that there is a ball with an infinite label on it - and by extension that the ball is the ωth ball added, where ω is an infinite ordinal. It is not possible for this ball to be associated with a finite step, since then the ball would have a finite label and the ordinal would be finite as well. It follows that such a ball "bω" could not have been added at any point before midnight, since at any point before midnight we are still working only with finite steps and finite balls. So it must have been added at or after midnight, which I will refer to as "step ω" since it would have to be the ωth step.

Let's examine what is in the set at the beginning of step ω, and call a ball in that set bx. If x is finite, then bx would have been removed at tx, which is prior to midnight. It's midnight right now, so bx is not in the jug. So x is not finite. If x is infinite, then it must have been added at or after midnight, but it's midnight right now and we haven't added any balls yet. So x is not infinite. If x is neither finite nor infinite, then it does not exist.

To illustrate this, let's perform the supertask twice. The first time we do it, let's paint every ball removed blue, and every ball remaining in the jug red. So we have a blue ball for every natural number, and a red ball for every ball in the set of balls never removed (infinite numbers or unknown numbers or whatever). Dump out the red balls and start the supertask again.

You will find that there is never a time when there is both a red and a blue ball in the jug.You will find that all blue balls are added before midnight, and no red ball is added before midnight.You will find that all blue balls are removed before midnight, and no red ball is removed before midnight.

The only way, then, to end up with red balls in the jug is to have an empty jug at midnight, and then to add red balls at or after midnight. And since the puzzle does not prescribe adding anything at all at or after midnight, there is no reason for us to do anything to the empty jug that we have at midnight. Then it was also empty the first time we did it, and there were no balls to paint red in the first place.

Now, I predict that you will make a fuss about how red balls get added "the same way" that infinite labels get added when we append a 0, or "the same way" X happens in some Y supertask that is different. That just isn't true. It's not "the same way." I don't think you even understand what the "way" is in the first place in order to claim that anything is "the same."

I wonder if this situation is not as impossible as it seems. Suppose there exist infinitely many numbered "balls" moving on disjoint paths. Define the "jug" to be the minimal region of space satisfying the conditions. Unbounded speed is not necessary; the conditions can be met if the higher numbered "balls" get progressively smaller, needing to travel less distance to enter and exit the "jug."

So if space and time are continuous there is no need for super speed, and in some sense this jug puzzle could be happening all the time.

PeteP wrote:What is the fundamental difference between the arguments that there can't be balls that were added in no step and that the end state can't be something that can't result from any step?

The argument isn't that there can't be balls that were added in no step, the argument is that we can follow the path of the individual balls so we *know* they aren't in the jug. We know where they are now, which is outside the jug.

Otherwise, it would certainly be possible that all the removed balls were back in the jug at midnight.

gmalivuk wrote:Your problem is your continued insistence that somehow balls like ω and ω+1 and 2ω just magically show up. All the balls we started with had natural numbers on them. So which natural number was originally on the now-lowest ball in the jug, which you claim is now numbered at least ω? When did it acquire its new label? How did this happen when all we've been doing is coloring some balls and removing balls one-by-one from the jug?

PeteP wrote:What is the fundamental difference between the arguments that there can't be balls that were added in no step and that the end state can't be something that can't result from any step?

Where each ball is at midnight can be deduced by following its path throughout the supertask, as Ermes Marana said.

All natural-numbered balls are put into the jug at some point and then taken out at some later point. Thus we can conclude that there are no natural-numbered balls in the jug at midnight.All other balls are never put into the jug in the first place, and so remain either outside of it or nonexistent. Thus we can conclude that there are no non-natural-numbered balls in the jug at midnight.

If you take issue with our use of the pointwise (ball-wise) limit, then I think you have to conclude that the limit simply doesn't exist. There's still no consistent way (as in consistent with the axioms of the set theory that gives us transfinite ordinals in the first place) to conclude that there are infinite-numbered balls in the jug at midnight.

Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.---If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

No, if I accepted that the task can be finished then it being empty follows from that imo which is why (along with other examples of infinity being weird) I am not quite convinced that stuff with reaching infinity (and not just going against infinity) is something that makes sense at all. But that is beside the point of my question. The point wise limit is about what is happening with balls that are added and tangential to the statement of it not being possible for there to be balls with labels that aren't finite. I don't think that it makes sense for there to be balls with labels that are added at no step. But I don't think it makes sense for the endstate to be a state that is produced by no step, either. And I don't see the fundamental difference between the two arguments so I am asking for it. What is the fundamental difference between (at midnight) there suddenly being balls that were never added and a state that was never reached?

PeteP wrote:I am not quite convinced that stuff with reaching infinity (and not just going against infinity) is something that makes sense at all.

So how do you feel about Zeno?

The point wise limit is about what is happening with balls that are added and tangential to the statement of it not being possible for there to be balls with labels that aren't finite.

It's not tangential at all. The pointwise limit of a ball that is never added to the jug must have that ball outside the jug. Since infinite-numbered balls are never added to the jug, they must not be in the jug at the end.

What is the fundamental difference between (at midnight) there suddenly being balls that were never added and a state that was never reached?

The fundamental difference is that the jug being empty can be concluded by following the path of every ball, and the jug containing infinite-numbered balls cannot be concluded by any such reasoning.

Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.---If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

PeteP wrote:What is the fundamental difference between (at midnight) there suddenly being balls that were never added and a state that was never reached?

I think this is a really good question, because in a way it highlights why it can be so difficult to adjust our intuitions to reflect the mathematics of the problem. We argue that there is never a step that adds an infinite labeled ball, but ignore the argument that there is never a step that reduces the ever-increasing cardinality of the set of balls. Why does one argument trump the other?

The answer is hard to articulate, but it comes down to there being a large conceptual difference between what a discrete element of a set is, and what the contents of a set is. Any ball is a discrete element of a set (either the set of balls in the jug or the set of balls outside of the jug). The contents of a set is the result of the whole of an infinite number of actions.

We can look at a very trivial example that illustrates this. At step n add ball bn to the jug. Obviously, the jug at midnight has the infinite set B = {bn | n∈N}. But this is an end state that was never "reached." In fact, even after you've added a Graham's Number of balls, more balls than you can possibly imagine, at a point so close to midnight that you can't even really tell the time apart... you still have an infinite way to go. But it's not controversial that we end up with an infinite number of balls in the jug. It is a result of the whole of the infinite number of additions.

There are also no balls bx such that x∉N, because there is no such defined step x.

So really, the end state of the jug is always going to be the result of an infinite number of actions which themselves never "reach" the end state. Then any argument about never reaching the end state is irrelevant. Even arguing that the end state is an infinite number of balls (disregarding what the labels actually are) is to defend an end state that can never be reached.

On the other hand, a ball being put in the jug is independent of those infinite actions. It is a discrete object that is, at any time t, is either in the jug or not in the jug, regardless of what other balls are in the jug or not. If we take any ball and assign to it a value of 0 when it is not in the jug, and 1 when it is, then to argue that an infinite number of steps can put it in the jug (when it is not in the jug at any finite step) is to suggest that the limit of the sequence 0, 0, 0, 0, ... is 1. And it is not.

So what it comes down to is this:The properties of the set of balls in the jug are the result of an infinite number of actions.The properties of an individual ball are not the result of an infinite number of actions.

gmalivuk wrote:Your problem is your continued insistence that somehow balls like ω and ω+1 and 2ω just magically show up. All the balls we started with had natural numbers on them. So which natural number was originally on the now-lowest ball in the jug, which you claim is now numbered at least ω? When did it acquire its new label? How did this happen when all we've been doing is coloring some balls and removing balls one-by-one from the jug?

No, the problem is in treating ω as a cardinal number. N is an infinite set. If you set aside the numbers 1 to 10, what's left is still an infinite set, isomorphic to ω. If you set aside the number 1 million then what's left is still an infinite set isomorphic to ω. If you contrive to set aside the last number, by some procedure such as ucim's transfer jug, then what's left is an infinite set, isomorphic to ω. That last number, in this particular arrangement, can be given the label ω. If you contrive to set aside the last 10 numbers then they can be labelled ω,ω+1,ω+2,...ω+9. They have the property that there are an infinite number of numbers smaller than them. If you contrive to set aside a countably infinite set from the top end, as is done by the 'remove-lowest' variant of the original puzzle, then you still leave behind a set ω, and the infinite set that is set aside is also isomorphic to ω, it is {ω,ω+1,ω+2,...}. The actual numbers have the property of having an infinite number of numbers smaller than them.

Xias wrote:Let's examine what is in the set at the beginning of step ω, and call a ball in that set bx. If x is finite, then bx would have been removed at tx, which is prior to midnight. It's midnight right now, so bx is not in the jug. So x is not finite. If x is infinite, then it must have been added at or after midnight, but it's midnight right now and we haven't added any balls yet. So x is not infinite. If x is neither finite nor infinite, then it does not exist.

At midnight, because of the construction of the supertask, it's not possible to declare which (non-finite) steps have or haven't happened.

Xias wrote:To illustrate this, let's perform the supertask twice. The first time we do it, let's paint every ball removed blue, and every ball remaining in the jug red. So we have a blue ball for every natural number, and a red ball for every ball in the set of balls never removed (infinite numbers or unknown numbers or whatever). Dump out the red balls and start the supertask again.

You will find that there is never a time when there is both a red and a blue ball in the jug.You will find that all blue balls are added before midnight, and no red ball is added before midnight.You will find that all blue balls are removed before midnight, and no red ball is removed before midnight.

The only way, then, to end up with red balls in the jug is to have an empty jug at midnight, and then to add red balls at or after midnight. And since the puzzle does not prescribe adding anything at all at or after midnight, there is no reason for us to do anything to the empty jug that we have at midnight. Then it was also empty the first time we did it, and there were no balls to paint red in the first place.

I'm not 100% sure I follow your argument, or understand the procedure(s) you are describing. But dividing the infinite input set up into two infinite sets, and then repeating the process on one of those infinite sets can produce two more infinite sets.

Take two Hilbert hotels, A and B, where A is occupied and B is empty. Tell everyone in A to write their current room number on a piece of paper and put it in their pocket. Now fill each room Bn by taking the occupant of A1 and moving everyone in A down one room. A will still be full at the end of the supertask, as will B - but what number is written on the piece of paper held by the current resident of A1?We have divided the original infinite set into two, one of which is numerically higher than the other. We could repeat the process, filling a new hotel C with the occupants of B, still leaving B full. And so on.

To be honest I'm coming to the view that the problem is poorly specified, in that it's of the form "What if this never-ending thing actually ended?" In some circumstance the answer can be presented using the same terms as the question - the 'remove highest' variation has the answer 'a never-ending set of things'. In other circumstances where the answer depends on the final stages of the never-ending thing, the answer depends on how much priority you give to the 'never-endingness' of the task versus the 'actually-endedness'.

However I find it strange that the argument that 'the jug sometimes ends up empty depending on how the balls are removed' does not square up with the observation that if the balls are unnumbered, or if no heed is paid to the numbering, that there is no way for the result to vary from one run of the experiment to the next. There will always be an infinite number of balls left in the jug.

The set-theory claim that the jug is empty after the 'remove-lowest' supertask is really a mathematical restatement of the observation that every finite ball that's added also gets removed, but whether you accept it or not, it does nothing to address the paradox - it simply restates it in mathematical terms.

If the balls are unnumbered then you'll always get an infinite set at the end. If they are numbered and you don't pay attention to the numbers, the same thing happens, with absolutely no way to end up with an empty jug. But if they are numbered and you happen to move them in a particular way, suddenly something different is predicted. That's the paradox which is not explained by the set-theory argument that the jug is empty. To resolve the paradox you need to explain how performing the task without caring about numbers could sometimes produce one result, and sometimes another. Either that or present a set-theory argument that always produces the same consistent result as the un-numbered result (which is my preferred approach.)

There is no such argument, because if you don't define how balls are removed then you can't deine the limit, and if you remove balls in a different way you get a different limit. If you don't like it, define your own idea of a limit and mathematicians can figure out if it's a useful definition.

The pointwise limit of a sequence of sets is the conventional way to take limits of sets, and that limit is unequivocally an empty jug at the end.

Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.---If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

You can, for steps 1->k, put a ball in the jug, and then for all following steps, put a ball in the jug and take one out. The jug will have k balls in it at the end if you leave those original k balls alone, but will have no balls in it if you remove the first k balls first. Each ball has a path. In the first case, you cannot guarantee that every ball gets removed (in fact, the first k balls do not get removed), but in the second case you can. This is what distinguishes this puzzle from the limit of 1+1-1+1-1.... In this case, the "1"s don't have a path; they are indistinguishable. But balls are distinguishable.

Spoiler:

I suppose if you did this with electrons, the result would be different.

kryptonaut wrote:If you contrive to set aside the last number, by some procedure such as ucim's transfer jug, then what's left is an infinite set, isomorphic to ω.

No, if you contrive to set aside the last number, you will find that there is no last number.

That's the key to all this. There is no last number. And thus, there is no number after the last number (which is the number ω you think ends up in the jug).

So, ω never ends up in the jug.

And since the "last number" never ends up in the jug, and every other number has been removed, the jug cannot be anything other than empty.

Jose

Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.